Question

$\displaystyle \int{{{{\sin }^3}\left( {\frac{2}{3}x} \right){{\cos }^4}\left( {\frac{2}{3}x} \right)\,dx}}$

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Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm \: \displaystyle \int{{{{\sin }^3}\left( {\frac{2}{3}x} \right){{\cos }^4}\left( {\frac{2}{3}x} \right)\,dx}}$

can be rewritten as

$\rm \: =  \:\displaystyle \int{{{{\sin }^2}\left( {\frac{2}{3}x} \right){{\cos }^4}\left( {\frac{2}{3}x} \right)sin\left( {\frac{2}{3}x} \right)\,dx}}$

can be further rewritten as

$\rm \: =  \:\displaystyle \int{{{\bigg(1 - {\cos}^2}\left( {\frac{2}{3}x} \right)\bigg){{\cos }^4}\left( {\frac{2}{3}x} \right)sin\left( {\frac{2}{3}x} \right)\,dx}}$

To evaluate this integral, we use method of Substitution.

So, Substitute

$\rm \: cos\left( {\dfrac{2}{3}x} \right) \: = \: y$

$\rm \: - \: \dfrac{2}{3} \: sin\left( {\dfrac{2}{3}x} \: \right) \: = \: \frac{dy}{dx}$

$\rm \: sin\left( {\dfrac{2}{3}x} \right)dx \: = \: - \: \dfrac{3}{2} \: dy \:$

So, on substituting the values, we get

$\rm \: =  \: - \: \dfrac{3}{2} \displaystyle \int\rm \: (1 - {y}^{2}) {y}^{4} \: dy$

$\rm \: =  \: - \: \dfrac{3}{2} \displaystyle \int\rm \: ( {y}^{4} - {y}^{6}) \: dy$

$\rm \: =  \: \dfrac{3}{2} \displaystyle \int\rm \: ( {y}^{6} - {y}^{4}) \: dy$

We know,

$\boxed{\sf{  \:\rm \: \displaystyle \int \: {x}^{n} \: dx \: = \: \frac{ {x}^{n + 1} }{n + 1} + c \: \: }}$

So, using this, we get

$\rm \: =  \:\dfrac{3}{2}\bigg(\dfrac{ {y}^{7} }{7} - \dfrac{ {y}^{5} }{5} \bigg) + c$

$\rm \: =  \:\dfrac{3}{14} {cos}^{7}\left( {\dfrac{2}{3}x} \right) - \dfrac{3}{10} {cos}^{5}\left( {\dfrac{2}{3}x} \right) + c$

Hence,

$\rm\implies \:\displaystyle \int{{{{\sin }^3}\left( {\frac{2}{3}x} \right){{\cos }^4}\left( {\frac{2}{3}x} \right)\,dx}} \\ \\ \rm \: =  \:\dfrac{3}{14} {cos}^{7}\left( {\dfrac{2}{3}x} \right) - \dfrac{3}{10} {cos}^{5}\left( {\dfrac{2}{3}x} \right) + c$

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Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$

Answer 2

$\sf I = \int sin^{3}(\dfrac{2x}{3}) cos^{4} (\dfrac{2x}{3})$

Now, we can write sin³(2x/3) as sin²(2x/3)sin(2x/3),

$\sf I = \int sin^{2}(\dfrac{2x}{3}) sin(\dfrac{2x}{3} ) cos^{4} (\dfrac{2x}{3})$

We now can use sin²x + cos²x = 1 and replace sin²(2x/3) by (1 - cos²(2x/3))

$\sf I = \int (1 - cos^{2}(\dfrac{2x}{3} ) sin (\dfrac{2x}{3} ) cos^{4} (\dfrac{2x}{3} )$

Now just to make the question look neat and allow us to think better, we will take cos(2x/3) as t and then differentiate,

$\sf cos(\dfrac{2x}{3}) = t$

$\sf - sin (\dfrac{2x}{3} ) \times \dfrac{2}{3}dx = dt$

We can clearly see that we have sin(2x/3)dx in our question,so we will express it in simpler manner so as to substitute its value in the question to make it easy to think ahead,

$\sf sin (\dfrac{2x}{3} )dx = -\dfrac{3}{2} dt$

Now all we are left with is substituting and doing some easy integration to reach to our destination,

$\sf I = \int - \dfrac{3}{2} (1 - t^{2} ) t^{4} dt$

[∵ cos (2x/3) = t => cos²(2x/3) = t²]

Now multiply the t⁴ with the entire bracket term,

$\sf I = -\dfrac{3}{2} \int (t^{4} - t^{6} ) dt$

Using the most basic yet most powerful integration formula : xⁿ = xⁿ⁺¹/n+1,we get this following result,

$\sf I = -\dfrac{3}{2} [\dfrac{t^{5} }{5} - \dfrac{t^{7} }{7} ]$

Now just put the values,

$\sf I = -\dfrac{3}{2} [\dfrac{cos^{5} (\dfrac{2x}{3}) }{5} - \dfrac{cos^{7} (\dfrac{2x}{3}) }{7} ] + c$

Finally multiply the -3/2 with the whole bracket,

$\sf I = -\dfrac{3}{10} cos^{5}(\dfrac{2x}{3}) + \dfrac{3}{14} cos^{7}( \dfrac{2x}{3}) + c$